Answer:
Step-by-step explanation:
mass of water to be heated = volume x density of water .
4000 x .9991 g
= 3.9964 kg
heat required per night = 3.9964 x 4.2 x ( 100 - 15 ) kJ
= 1426.71 kJ
calorific value of liquid butane = 50400 kJ / kg
so gram of liquid butane required = 1426.71 / 50400 kg
= .0283 kg
= 28.30 g
volume of liquid butane = mass / density
= 28.3 / .6
= 47.17 mL